Suppose X has an exponential distribution with lambda =3 . Determine the following. Round the answers to 3 decimal places.
a) P(X <= 0) = b) P(X >= 2) =
c) P(X <= 1) =
d) P(1 < X < 2) =
e) Find the value of x such that P(X < x) = 0.05 .
The correct answer and explanation is :
To solve the problem, we first need to recall the formula for the cumulative distribution function (CDF) of the exponential distribution. Given that the random variable $X$ follows an exponential distribution with parameter $\lambda = 3$, the probability density function (PDF) is:
$$
f(x) = \lambda e^{-\lambda x} \quad \text{for} \quad x \geq 0
$$
The cumulative distribution function (CDF) is:
$$
F(x) = 1 – e^{-\lambda x}
$$
Where $\lambda = 3$. Now, we will solve each part of the question step by step.
a) $P(X \leq 0)$
We can use the CDF to find $P(X \leq 0)$:
$$
P(X \leq 0) = F(0) = 1 – e^{-\lambda \cdot 0} = 1 – e^{0} = 1 – 1 = 0
$$
So, $P(X \leq 0) = 0$.
b) $P(X \geq 2)$
We can find $P(X \geq 2)$ by subtracting the CDF at $x = 2$ from 1:
$$
P(X \geq 2) = 1 – F(2) = 1 – (1 – e^{-\lambda \cdot 2}) = e^{-3 \cdot 2} = e^{-6} \approx 0.002478
$$
So, $P(X \geq 2) \approx 0.002$.
c) $P(X \leq 1)$
Using the CDF at $x = 1$:
$$
P(X \leq 1) = F(1) = 1 – e^{-\lambda \cdot 1} = 1 – e^{-3 \cdot 1} = 1 – e^{-3} \approx 0.9502
$$
So, $P(X \leq 1) \approx 0.950$.
d) $P(1 < X < 2)$
To find $P(1 < X < 2)$, we subtract the CDF at 1 from the CDF at 2:
$$
P(1 < X < 2) = F(2) – F(1) = (1 – e^{-6}) – (1 – e^{-3}) = e^{-3} – e^{-6}
$$
Approximating the values:
$$
e^{-3} \approx 0.0498, \quad e^{-6} \approx 0.002478
$$
$$
P(1 < X < 2) \approx 0.0498 – 0.002478 = 0.047322
$$
So, $P(1 < X < 2) \approx 0.047$.
e) Find the value of $x$ such that $P(X < x) = 0.05$
We need to solve the equation $F(x) = 0.05$, where $F(x) = 1 – e^{-3x}$. Set this equal to 0.05:
$$
1 – e^{-3x} = 0.05
$$
$$
e^{-3x} = 0.95
$$
Taking the natural logarithm of both sides:
$$
-3x = \ln(0.95)
$$
$$
x = \frac{-\ln(0.95)}{3} \approx \frac{0.051293}{3} \approx 0.0171
$$
So, $x \approx 0.017$.
Final Answers:
- a) $P(X \leq 0) = 0$
- b) $P(X \geq 2) \approx 0.002$
- c) $P(X \leq 1) \approx 0.950$
- d) $P(1 < X < 2) \approx 0.047$
- e) $x \approx 0.017$
Explanation:
The exponential distribution models the time between events in a Poisson process, where $\lambda$ is the rate parameter that determines the frequency of events. In this case, $\lambda = 3$, which means the expected time between events is $\frac{1}{\lambda} = \frac{1}{3} \approx 0.333$ units of time. The CDF, $F(x) = 1 – e^{-\lambda x}$, gives the probability that $X$ is less than or equal to a specific value. By plugging different values of $x$ into the CDF formula, we can compute the required probabilities.